Harold Vance Department of Petroleum Engineering
Texas A&M University
Petroleum Engineering 201S
Introduction to Petroleum Engineering
Notes on Dimensions and Units
Spring 2002
2
Fluid Properties
1.
Liquid Specific Gravity
γ liquid =
2.
141.5
γ liquid
− 131.5
Gas Specific Gravity
γ gas =
4.
ρ water
API Gravity
° API =
3.
ρ liquid
ρ gas
ρ air
=
Mg
M air
=
M g(pure gas)
28.97
=
M apparent( gas mixture)
28.97
Gas in Solution
Rs = standard cubic feet of gas liberated when one stock tank barrel of crude oil is
produced
5.
Oil Formation Volume Factor
B=
volume occupied at reservoir pressure and temperature
volume occupied at standard pressure and temperature
volume of oil + dissolved gas at reservoir conditions
Bo = volume of oil entering stock tank at standard conditions . In other words:
Bo =
reservoir barrels of oil & dissolved gas  RB 


stock tank barrel of oil
 STB 
mass of oil mass of gas
+
ρ STO + 0.01357 R s γ g
STB
STB
=
=
,
mass of reservoir liquid
ρ oR
RB
in
where: ρSTO and ρoR are densities of the stock tank oil and reservoir liquid, both
3
lbm/ft , and 0.01357 converts the gas volume to mass.
The conversion factors is derived as follows:
 Scf  lb ⋅ mole
Rs

 STB  380.7 Scf
 28.97 γ g lb m


 lb ⋅ mole
 STB 
lb

 = 0.01357Rs γ g m .


3
ft 3
 5.615ft 
3
Example Fluid Properties Calculations
1.
Liquid Specific Gravity
Given
3
Density of Liquid
Density of Water
Definition
γ liquid =
Calculation
γ liquid =
2.
48.6 lbm/ft
3
62.4 lbm/ft
ρ liquid
ρ water
ρ liquid
ρ water
=
48.6
= 0.779
62.4
API Gravity
Given
Specific gravity of liquid-1
Specific gravity of liquid-2
Specific gravity of liquid-3
0.779
0.876
1.000
Definition
° API =
141.5
− 131.5
γ liquid
Calculations
Specific Gravity
0.779
0.876
1.000
3.
API Gravity
50.0
30.0
10.0
Gas Gravity
Given
Component
Composition (Mole fraction)
Methane C1
Ethane C2
Propane C3
n-Butane n-C4
Definition
γ gas =
ρ gas
ρ air
=
Mg
M air
=
0.850
0.090
0.040
0.020
M g(pure gas)
28.97
4
=
M apparent( gas mixture)
28.97
3.
Gas Gravity (continued)
Calculation
Component
j
Mole fraction
yj
Molar Mass
Mj
C1
C2
C3
n-C4
0.850
0.090
0.040
0.020
1.000
16.04
30.07
44.10
58.12
γ gas =
4.
Component
Mass
yjM j
13.63
2.71
1.76
1.16
19.26
Mass fraction
y j M j / 19.26
0.708
0.141
0.091
0.060
1.000
M apparent( gas mixture) 19.26
= 0.665
28.97
28.97
Oil Formation Volume Factor
Given
3
Density of Reservoir Oil
Density of Stock Tank Oil
Gas in Solution
Gas Gravity
Definition
Bo =
Calculation
Bo =
5.
47.5 lbm/ft
3
55.5 lbm/ft
400 Scf/STB
0.72
ρ STO + 0.01357 R s γ g
ρ oR
55.5 + (0.01357)(400)(0.72)
RB
= 1.251
47.5
STB
Gas Formation Volume Factor
Given
Reservoir Pressure
1000 psi
o
Reservoir Temperature
610 R
Gas Law Deviation Factor (z) 0.90
Standard Pressure
14.65 psi
o
Standard Temperature 60 F
Definition
Bg =
Calculation
Bg =
Vrc
Vsc
Vrc (nzRT / p )rc (0.9)(610)(14.65)
Rcf
RB
=
=
= 0.0155
= 0.00275
(nRT / p )sc (60 + 460)(1000)
Vsc
Scf
Scf
5
Fluid Properties Nomenclature
Lower Case Letters
n
Amount, moles
Amount at reservoir conditions, moles
nrc
nsc
Amount at surface conditions, moles
p
Pressure, FL-2
Pressure at reservoir conditions, FL-2 (psia)
prc
Pressure at surface conditions, FL-2 (psia)
psc
z
Gas law deviation factor, actual volume/ideal volume
zrc
Gas law deviation factor at reservoir conditions, actual volume/ideal volume
zsc
Gas law deviation factor at surface conditions, actual volume/ideal volume (1.0)
Upper Case Letters
B
Formation volume factor, reservoir volume/surface volume
Bg
Gas formation volume factor, reservoir volume/surface volume (rcf/scf)
Oil formation volume factor, reservoir volume/surface volume (rb/stb)
Bo
Water formation volume factor, reservoir volume/surface volume (rb/stb)
Bw
M
Molar mass, mass/mole
Apparent molar mass of a gas mixture, mass/mole (lbm/lb·mole)
Ma
Mair Molar mass of air, mass/mole (28.97 lbm/lb·mole)
Molar mass of gas, mass/mole (lbm/lb·mole)
Mg
R
Gas-oil ratio, surface gas volume/surface oil volume
Produced gas-oil ratio, surface gas volume/surface oil volume (scf/stb)
Rp
Solution gas-oil ratio, surface gas volume/surface oil volume (scf/stb)
Rs
T
Temperature
V
Volume, L3
Volume at reservoir conditions, L3 (ft3)
Vrc
Volume at surface conditions, L3 (ft3)
Vsc
Greek Letters
γl
Liquid specific gravity, density of liquid at 60oF/density of water at 60oF
ρ
Mass density, mass/volume
Density, mass/volume (lbm/ft3)
ρg
Liquid density, mass/volume (lbm/ft3)
ρl
Oil density, mass/volume (lbm/ft3)
ρo
Oil density at reservoir conditions, mass/volume (lbm/ft3)
ρoR
ρSTO Oil density at surface (stock tank) conditions, mass/volume (lbm/ft3)
Water density, mass/volume (lbm/ft3)
ρw
6
Dimension and Unit Systems
Dimensions are physical quantities and units are standards of measurement. Thus, dimensions
are independent of units.
Dimensions are classified as fundamental, supplementary, and derived. Supplementary
dimensions can be considered as fundamental dimensions. Fundamental dimensions are those
necessary to describe a particular field of engineering or science. Derived dimensions are
combinations of fundamental dimensions. A dimensional system is just the smallest number of
fundamental dimensions to form a consistent and complete set for a field of engineering or
science. A dimensional system is called an absolute system if its dimensions are not affected by
gravity; otherwise it is called a gravitational system.
Units are classified as base, supplementary, and derived. Units for supplementary dimensions
can also be considered as base units. Derived units are combinations of base units.
A dimension and unit system is a set of fundamental dimensions and base units necessary for a
particular field of science or engineering. It is called a coherent system if equations between
numerical values (units) have the same form as the corresponding equations between the
quantities (dimensions). For example, in the SI system, F = m·a is used to define the derived
dimension. Likewise, 1 newton = (1 kilogram)·(1 meter per second squared) is used to define
the derived unit. In other words, in a coherent system, combinations of any two unit quantities is
the unit of the resulting quantity. Coherency is a major advantage of the SI system.
In petroleum engineering, three dimension and unit systems are commonly used:
•
•
•
The International System of Units (SI Units)
The American Engineering System of Units (Oilfield Units)
The Darcy System of Units (Darcy Units)
The purpose of these notes is to help you learn the above systems and how to convert units from
one system to another.
7
The International System of Units (SI)
Fundamental Dimension
Base Unit
length [L]
mass [M]
time [t]
electric current [I]
absolute temperature [T]
luminous intensity [l]
amount of substance [n]
meter (m)
kilogram (kg)
second (s)
ampere (A)
kelvin (K)
candela (cd)
mole (mol)
Supplementary Dimension
Base Unit
plane angle [θ]
solid angle [ω]
radian (rad)
steradian (sr)
Derived Dimension
Unit
Definition
acceleration [L/t2]
area [L2]
Celsius temperature [T ]
concentration [n/L3]
density [M/L3]
electric charge [It]
electric potential [ML2/It3]
electric resistance [ML2/I2t3]
energy [ML2/t2]
force [ML/t2]
frequency [1/t]
molar mass [M/n ]
power [ML2/t3]
pressure [M/Lt2]
quantity of heat [ML2/t2]
specific heat [L2/t2T]
thermal conductivity [ML/t3T]
velocity [L/t]
viscosity, dynamic [M/Lt]
volume [L3]
work [ML2/t2]
meter per second squared
square meter
degree Celsius (oC)
mole per cubic meter
kilogram per cubic meter
coulomb (C)
volt (V)
ohm (Ω)
joule (J)
newton (N)
hertz (Hz)
kilogram per mole
watt (W)
pascal (Pa)
joule (J)
joule per kilogram kelvin
watt per meter kelvin
meter per second
pascal second
cubic meter
joule (J)
m/s
2
m
K
3
mol/m
3
kg/m
A·s
W/A
V/A
N·m
2
kg·m/s
1/s
kg/mol
J/s
2
N/m
N·m
J/(kg·K)
W/(m·K)
m/s
Pa·s
3
m
N·m
8
2
The International System of Units (SI) (Cont'd)
Prefix
Decimal
Multiplier
atto
femto
pico
nano
micro
milli
centi
deci
deka
hecto
kilo
mega
giga
tera
peta
exa
10
-15
10
-12
10
-9
10
-6
10
-3
10
-2
10
-1
10
+1
10
+2
10
+3
10
+6
10
+9
10
+12
10
+15
10
+18
10
-18
Symbol
a
f
p
n
µ
m
c
d
da
h
k
M
G
T
P
E
9
American Engineering System of Units (AES)
Fundamental Dimension
Base Unit
length [L]
mass [M]
force [F]
time [t]
electric charge[Q]
absolute temperature [T]
luminous intensity [l]
amount of substance [n]
foot (ft)
pound mass (lbm)
pound force (lbf)
second (sec)
coulomb (C)
Rankine (oR)
candela (cd)
mole (mol)
Supplementary Dimension
Base Unit
plane angle [θ]
solid angle [ω]
radian (rad)
steradian (sr)
Derived Dimension
Unit
Definition
acceleration [L/t2]
area [L2]
Fahrenheit temperature [T ]
concentration [n/L3]
density [M/L3]
electric current [Q/t]
electric potential [FL/Q]
electric resistance [FLt/Q2]
energy [FL]
frequency [1/t]
molar mass [M/n ]
power [FL/t]
pressure [F/L2]
quantity of heat [FL]
velocity [L/t]
viscosity, dynamic [Ft/L2]
volume [L3]
work [FL]
foot per second squared
square foot
degree Fahrenheit (oF)
mole per cubic foot
pound mass per cubic foot
ampere (A)
volt (V)
ohm (Ω)
foot pound force
hertz (Hz)
pound mass per mole
foot pound force per second
pound force per square foot (psf)
british thermal unit (BTU)
foot per second
pound force second per square foot
cubic foot
foot pound force
ft/sec
2
ft
o
R-459.67
3
mol/ft
3
lbm/ft
C/sec
W/A
V/A
ft· lbf
1/sec
lbm/mol
ft· lbf/sec
2
lbf/ft
777.65 ft· lbf
ft/sec
2
lbf·sec/ft
3
ft
ft· lbf
10
2
Oilfield Units (related to AES System)
Dimension
Unit
Definition
area [L2]
energy [FL]
acre (ac)
horsepower hour (hp·hr)
kilowatt hour (kW·hr)
inch (in)
yard (yd)
mile (mi)
ounce (oz)
ton
horsepower (hp)
watt (W)
pound force per square inch (psi)
atmosphere (atm)
minute (min)
hour (hr)
day
centipoise (cp)
gallon (gal)
barrel (bbl)
acre·ft (ac·ft)
43,560 ft
6
1.98000 x 10 ft· lbf
6
2.6552 x 10 ft· lbf
1/12 ft
3 ft
5,280 ft
1/16 lbm
2000 lbm
550 ft· lbf/sec
0.73756 ft· lbf/sec
2
144 lbf/ft
14.696 psi
60 sec
3,600 sec
86,400 sec
-2
2
10 dyne·s/cm
3
0.133681 ft
3
5.614583 ft
3
43,560 ft
length [L]
mass [M]
power [FL/t]
pressure [F/L2]
time [t]
viscosity, dynamic [Ft/L2]
volume [L3]
11
2
Conversion of Units
In a coherent system of units such as SI, a derived dimension is a product or quotient of other
dimensions. For example, the dimension—force—is the product of mass and acceleration,
F = ma, and the unit of force—newton—is the product of the unit of mass and the unit of
acceleration. In the American Engineering System, force is expressed in lbf, mass in lbm, and
2
acceleration in ft/sec . This system is obviously not coherent. Hence, a conversion factor other
lb m ⋅ ft
ma
than one must be used in the equation; that is, F =
, where gc = 32.174
is a
2
gc
sec ⋅ lb f
constant, known as the gravitational conversion constant.
Derivation of gc
The principle of conservation of units is used to derive gc. Briefly stated, this principle is that to
convert a relationship (an equation) from a given system of units to another (a required) system
of units, the given units must be conserved. The technique is illustrated below.
We wish to convert F = ma from SI to AES. Here, SI is the given system and AES is the
required system.
First, find the dimension of the “hidden constant” in the given equation. In other words,
1=
F
ma
has dimension force divided by the product of mass and acceleration.
Next, consider the units of the hidden constant—the given units are
N⋅s
2
kg ⋅ m
and the required units are
lb f ⋅ sec
2
lb m ⋅ ft
.
Now, convert the given units of the hidden constant to the required units.
Thus,
1 lb f ⋅ sec
 1 N ⋅ s 2  0.45359 kg  0.3048 m  lb f 
=





 kg ⋅ m  lb m
 ft  4.4482 N  32.174 lb m ⋅ ft
12
2
Review Problems
1. A weir is a regular obstruction in an open flow
channel over which flow takes place such as that
shown in the sketch at right. It can be used to
measure open channel flow rates. For a rectangular
weir, the theoretical formula for the flow rate is
1.5
Q = 5.35LH
3
where Q is discharge rate in ft /sec,
L is length of the weir in ft,
H is height of fluid above the crest in ft.
Determine a new constant so the formula can be
3
applied with Q in m /s and L and H in m.
2. The ideal gas equation can be written,
pv = RT
where p = pressure, Pa
3
v = molar volume, m /(kg·mol)
3
R = gas constant, 8314.5 Pa·m /(kg·mol·K)
T = absolute temperature, K
2
3
Determine a new constant so that the equation can be applied with p in lbf/in , v in ft /(lbm·mol),
o
and T in R.
3. The universal law of gravity may be written,
-11 m m
F =  6.672 x 10  1 2 2

 r
where F = force of attraction between two bodies, N
m1 = mass of body one, kg
m2 = mass of body two, kg
r = distance between bodies, m
Determine a new constant so that the law can be applied with F in lbf, m1 & m2 in lbm, and r in
mi.
Answers
0.5
1. 2.95 m /s
3
o
2. 10.732 psi· ft /(lbm·mol· R)
-18
2
2
3. 1.192 x 10 lbf·mi /lbm
13
Porosity, Permeability, and Saturation (φ-k-S)
Porosity is a measure of the fluid storage capacity of a rock,
Vp
φ=
Vb
where φ = porosity, fraction
Vb = bulk volume = Vp + Vm
Vp = pore volume
Vm = matrix volume
Permeability is a measure of the fluid
conductivity of a rock. It is defined by
Darcy’s law, which is based upon experimental data. For horizontal, linear
flow of a liquid completely saturating
the rock,
k=
where
qµL
A∆p
k = permeability, d
q = flow rate, cm3/s
µ = fluid viscosity, cp
L = length of flow path, cm
A = cross-sectional area of
flow path, cm2
∆p = pressure difference across
flow path, atm
The dimension of permeability is [L2].
 qµL   L3 P ⋅ t L 1 1 
 A∆p  =  t ⋅ 1 ⋅ 1 ⋅ 2 ⋅ P 
L
 


Saturation is a measure the amount and type of fluid stored in a rock.
Sl =
Vl
Vp
where Sl = saturation of fluid l (oil, water, or gas)
Vl = fluid volume ( l = oil, water, or gas)
Vp = pore volume
14
Darcy’s Law
Darcy found that when water flows
vertically downward through sand, the
volume of water passing through the
system in unit time (the discharge q in
Fig. 2) is proportional to the drop in
head h across the sand. Considering
the cross sectional area A and thickness of the sand l, these observations
can be written,
v=
dh
q
= −K
A
dl
where K is a coefficient depending
upon the permeability of the sand and
the properties of the fluid.
Darcy’s original experiment has been
extended to answer questions about
flow of different fluids in porous
media of various permeabilities along
flow paths in various directions.
The generalized two dimensional form of Darcy’s law (API Code 27) is
ρg dz
k  dp
−6 
−
vs = − 
x 10 
µ  ds 1.01325 ds

where s = distance in direction of flow (always positive), cm
vs = volume flux across a unit area of the porous medium in unit time (q/A)
along flow path s, cm/s
z = vertical coordinate (positive downward), cm
3
ρ = density of the fluid, g/cm
2
g = acceleration of gravity, 980.665 cm/s
dp/ds = pressure gradient along path s at the point to which vs refers, atm/cm
µ = viscosity of the fluid, cp
k = permeability of the porous medium, d
15
The Darcy System of Units
Fundamental Dimension
Base Unit
length [L]
mass [M]
time [t]
centimeter (cm)
gram (g)
second (s)
Derived Dimension
Unit
Definition
acceleration [L/t2]
area [L2]
density [M/L3]
energy [ML2/t2]
force [ML/t2]
permeability [L2]
pressure [M/Lt2]
velocity [L/t]
viscosity, dynamic [M/Lt]
volume [L3]
work [ML2/t2]
centimeter per second squared
square centimeter
gram per cubic centimeter
erg
dyne
darcy (d)
atmosphere (atm)
centimeter per second
centipoise (cp)
cubic centimeter
erg
cm/s
2
cm
3
g/cm
dyne·cm
2
g·cm/s
2
cp·cm /atm·s
6
2
1.013250 x 10 dyne/cm
cm/s
-2
2
10 dyne·s/cm
3
cm
dyne·cm
In SI Units
Derived Dimension
Unit
permeability [L2]
square meter (m )
2
In Oilfield Units
Derived Dimension
Unit
permeability [L2]
millidarcy (md)
16
2
Converting Permeability Units
The principle of conservation of units is used. Briefly stated, this principle is that to convert a
relationship (an equation) from a given system of units to another (a required) system of units,
the given units must be conserved. The technique is illustrated below.
We wish to convert k =
qµL
from Darcy units (d, cm, s, cp, atm) to Oilfield units (md, ft, bbl,
A∆p
day, cp, psi). Here, Darcy units is the given system and Oilfield units is the required system.
First, find the dimension of the “hidden constant” in the given equation. In other words,
kA∆p
1=
has dimension kA∆p divided by qµL [1].
qµL
Next, consider the units of the hidden constant—the given units are
d ⋅ atm ⋅ s
md ⋅ ft ⋅ psi ⋅ day
and the required units are
.
2
cp ⋅ cm
cp ⋅ bbl
Now, convert the given units of the hidden constant to the required units.
Thus,
2
3
2
 1 d ⋅ atm ⋅ s  10 md  ft  14.696 psi  day  (30.48) cm 






2 
2

 
=
ft
 cp ⋅ cm  d  ft  atm  86,400 s 

3
 158.02 md ⋅ ft ⋅ psi ⋅ day  5.614583 ft  887.22 md ⋅ ft ⋅ psi ⋅ day
 =



cp ⋅ ft 3
bbl
cp ⋅ bbl



Converting from Darcy units to other systems of units is similar.
Summary of Darcy equations for horizontal, linear flow (including conversion factors):
Darcy Units
qµL
k=
A∆p
SI Units
qµL
k=
A∆p
Oilfield Units
qµL
k = 887.2
A∆p
2
12
2
NOTE: A m is a huge permeability unit! There are 1.01325 x 10 d/m .
2
Thus, the SPE preferred SI unit for permeability is a µm , introducing
12
a factor of 10 in the equation.
17
Horizontal, Linear Flow
A
q
q
L
x2
x1
Applying Darcy’s law,
vx =
k dp
q
=−
A
µ dx
Separate the variables and integrate to obtain the flow equation,
x2
kA p 2
q ∫ dx = −
∫ dp
⇒ q (cm 3 / s) = −
µ p
1
x1
kA(p 2 − p1 ) kA(p1 − p 2 ) kA∆p
=
=
µ(x 2 − x 1 )
µ (x 2 − x 1 )
µL
In oilfield units, the equation is
1.1271 x 10 −3 kA∆p
q(RB / day) =
µL
1.1271 x 10 −3 kA∆p
or q(STB / day) =
µBL
2
EXAMPLE: A rock sample10 cm long and 2 cm in cross section is used for some steady-state
flow tests. Calculate the permeability of the rock if it is completely saturated with
an oil having a viscosity of 2.5 cp and oil is flowed through the rock at a rate of
3
0.0080 cm /s under a 1.5 atm pressure drop?
(
)
0.0080 cm 3 / s (2.5 cp )(10 cm )
qµL
=
= 0.067 d = 67 md
k=
A∆p
2 cm 2 (1.5 atm )
(
)
18
Wellbore Volumes
Wellbores are right circular cylinders,
the volume of which is,
2
2
V = πr h = π d h
4
h
Many drilling and completion applications
require the annular volume,
(
V = π r22
2
− r1
)h = π4 (d 22 − d12 )h
EXAMPLE: Suppose you are the drilling engineer on a rig. You have set 7-5/8 inch OD,
33.7 lbm/ft casing (ID = 6.765 inches) from 0 to 5900 ft and have drilled out to
7900 ft. The average bore hole diameter from 5900 to 7900 ft is 6.25 inches.
3
Calculate the annular volume in ft if 4-1/2 inch OD casing is installed.

2
2



2
2



 6.25 in   4.5 in 
  6.765 in   4.5 in 

  2000 ft 
 − 
 5900 ft + 
 − 
V = π 
4  12 in / ft 
 12 in / ft   12 in / ft  
 12 in / ft  



V = 987 ft
3
19
Pressure Gradients
The basic equation of hydrostatics can be written
∆p = ρg∆h
where ∆p = change in fluid pressure, Pa
3
ρ = density, kg/m
2
g = acceleration of gravity, m/s
∆h = change in height of fluid, m
The pressure gradient (Pa/m) is
∆p
∆h
= ρg
There is no conversion factor in SI because
 Pa kg m N 1 Pa 
 m = m 3 ⋅ s 2 = m 2 ⋅ m = m 
2
In oilfield units, ∆p = change in fluid pressure, lbf/in or psi
3
ρ = density, lbm/ft
2
g = acceleration of gravity, ft/s
∆h = change in height of fluid, ft
and the pressure gradient (psi/ft) is
∆p
∆h
=
ρg
144g c
,where g c = 32.174
lb m ⋅ ft
lb f ⋅ s
2
Notice that the units check because
 psi lbm ft ft 2 lb f ⋅ s 2 lb f 1 psi 
 ft = ft 3 ⋅ s 2 ⋅ in 2 ⋅ lb ⋅ ft = in 2 ⋅ ft = ft 


f
3
EXAMPLE: Calculate the pressure gradient for a fluid having a density of 65 lbm/ft .
2
Assume g = 32 ft/s .
∆p
∆h
=
ρg
144g c
=
(65)(32) = 0.449 or 0.45 psi / ft
(144)(32.174)
20
Horizontal, Radial Flow
ds = -dr
r
h
re
rw
Applying Darcy’s law,
vr =
k dp
q
q
=
=
A 2 πrh µ dr
Separate the variables and integrate to obtain the flow equation,
re dr 2πkh p e
2πkh (p e − p w )
2πkh∆p
3
=
=
q ∫
∫ dp ⇒ q (cm / s) =
µ p
µ ln re r w
µ ln re r w
rw r
w
(
)
(
)
In oilfield units, the equation for liquid is
q l (RB / day) =
7.0819 x 10 −3 kh∆p
µ ln re r w
(
)
or q l (STB / day) =
and, the equation for gas is
q g (RB / day) =
7.0819 x 10 −3 kh∆p
µ ln re r w
(
)
or q g (Scf / day) =
21
7.0819 x 10 −3 kh∆p
µB l ln re r w
(
7.0819 x 10 −3 kh∆p
µB g ln re r w
(
)
)
Reservoir Volumes
h
re
rw
The volume occupied by the reservoir rock or bulk volume is
2
Vb = πre h
Since porosity is the ratio of the pore volume to bulk volume, the pore volume is
2
Vp = πre hφ
And, since water saturation is the fraction of the pore volume filled with water, the hydrocarbon
pore volume is
(
2
Vhc = πre hφ 1− S w
)
A commonly used equation to calculate the surface volume of oil in place is
N(STB) =
7758Ahφ(1 − S w )
Bo
, where A is the drainage area in acres.
and the surface volume of gas in place is
G(Scf) =
7758Ahφ(1 − S w )
Bg
, where B g is the gas formation volume factor in RB / Scf .
22
Homework Problems
1. A certain oilfield has an average producing gas-oil ratio of 350 scf/stb. The density of the stock
tank oil is 54.7 lbm/ft3 at 60 oF and the apparent molar mass of the associated gas is 21.7 lbm/lb⋅mole.
Determine the
a. API gravity of the stock tank oil.
b. Specific gravity of the associated gas.
c. Density of the reservoir oil if the oil formation volume factor is 1.2 rb/stb and the gas in
solution is equal to the average producing gas-oil ratio.
d. Type of reservoir fluid.
2. The equation of state for a particular substance at low pressures can be written
p = 53.55ρT,
where
p = pressure, lbf/ft2
ρ = density, lbm/ft3
T = absolute temperature, oR
a. Determine a new constant so the equation can be used with p in Pa, ρ in kg/m3, and T in K.
b. Use the ideal gas law (pV = nRT), the definition of quantity (mass/molar mass), and the
definition of density (mass/volume) to identify the substance.
3. Power is defined as work per unit time.
a. Find the units of the hidden constant in P = W/t (SI system), where P is power in
watts, W is work in newton-meters, and t is in seconds.
b. Convert to constant to the AES system, where P is power in horsepower, W is work in
foot-pounds of force, and t is in minutes.
4. A cylindrical sandstone sample, one inch in diameter and one inch long has a pore volume of 0.275
in3. The pores are partially filled with 0.135 in3 of water. What is the porosity and water saturation of the
sample?
5. A cylindrical sandstone sample, one inch in diameter and one inch long is placed in a permeameter and
water having a viscosity of one centipoise is flowed through the circular faces of the sample at a rate of
one cm3/s under a pressure difference of one atmosphere. What is the permeability of the sample?
6. An acre-ft of reservoir has a bulk volume of 43,560 ft3. Calculate the volume of water (in barrels per
acre-ft) for a sandstone reservoir having a porosity of 20 % and a water saturation of 25 %.
7. A rock sample is 1 inch long and 1 inch in diameter. The sample is used for some steady-state flow
tests. Calculate the permeability of the rock if it is completely saturated with water having a viscosity of
1 cp and water is flowed through the sample at a rate of 0.5 in3/min and the pressure drop is 10 psi.
8. Find the conversion constant to calculate permeability in md from a horizontal, linear flow experiment
when flow rate is in cm3/s, viscosity is in cp, length is in cm, and pressure drop is in dynes/cm2.
9. Find the conversion constant to calculate permeability in md from a horizontal, linear flow experiment
when flow rate is in m3/day, viscosity is in cp, length is in m, and pressure drop is in kPa.
23
10. Suppose you are the drilling engineer on a rig. You have set 9 5/8 in OD, 40 lbm/ft casing (ID =
8.835 in) from 0 to 2000 ft and have drilled a 7 7/8 in bore hole from 2000 to 4000 ft.
a. Find the capacity of the hole in barrels from 0 to 4000 feet.
b. The hydrostatic pressure at 4000 ft with fresh water in hole.
c. The annular volume if 4.5 inch OD, 11.60 lbm/ft casing is installed from 0 to 4000 ft
with fresh water in hole.
11. Find the conversion constant to calculate pressure gradient in psi/ft when fluid density is in lbm/gal
(ppg)
12. A vertical gas well is 6000 ft deep.
a. What is its bottom hole temperature if the mean surface temperature is 60 oF and the
local temperature gradient is 1.1 oF/100 ft?
b. What is its bottom hole flowing pressure if the surface flowing pressure is 900 psi
and the pressure gradient in the well bore is 0.15 psi/ft?
13. Assume a vertical well drains a right cylindrically shaped oil reservoir for which you have the
following data:
drainage area
formation thickness
radius of well bore
porosity
water saturation
permeability
oil viscosity
pressure at external radius
pressure at the well bore radius
40 acres (external radius = 745 ft)
25 feet
0.25 feet
19 %
21 %
75 millidarcy
2.1 centipoise
3545 psi
1515 psi
Calculate the initial production rate in reservoir barrels per day.
14. Given the following data on a gas reservoir:
porosity
water saturation
gas formation volume factor
0.22
0.30
1.0 rb/MScf
a. Calculate the volume of gas in MScf per acre-foot of reservoir volume.
b. If the recovery factor is 80 %, what is the recoverable reserve of a 160 acre spaced well
with an average reservoir thickness of 15 feet?
24
Problem Layout
25
Conversion Factors
To convert from
acres
acres
acre-feet
atmospheres
atmospheres
atmospheres
atmospheres
barrels(oil)
barrels(oil)
barrels(oil)/d
Btu
Btu
Btu
Btu
Btu
centimeters
centimeters
centimeters
cm of mercury
cm of mercury
cm of mercury
cm of mercury
cm/s
cm/s
cm/s
cm/s
centipoise
cubic cm
cubic cm
cubic cm
cubic cm
cubic cm/s
cubic feet
cubic feet
cubic feet
cubic feet
cubic ft/d
cubic feet/h
cubic feet/min
cubic feet/s
cubic meters
cubic meters
Multiply by
4.356 x 104
4.047 x 103
4.356 x 104
7.6 x 101
3.39 x 101
1.01325 x 105
1.4696 x 101
5.614583
4.2 x 101
2.295 x 10-3
1.0550 x 1010
7.7816 x 102
3.927 x 10-4
1.055 x 103
2.928 x 10-4
3.281 x 10-2
3.937 x 10-1
6.214 x 10-6
1.316 x 10-2
4.461 x 10-1
1.333 x 103
1.934 x 10-1
1.969
3.281 x 10-1
3.6 x 10-2
2.237 x 10-2
1.0 x 10-3
6.28982 x 10-6
3.531 x 10-5
2.642 x 10-4
1.0 x 10-3
5.434 x 10-1
1.781 x 10-1
2.832 x 10-2
7.48052
2.832 x 101
3.278 x 10-7
4.275
4.720 x 10-1
4.48831 x 102
6.28982
3.531 x 101
26
To obtain
sq feet
sq meters
cubic feet
cm of mercury at 0 oC
ft of water at 4 oC
pascals
pounds-force/sq in
cubic feet
gallons
cubic meters/s
ergs
ft-pounds
horsepower-hours
joules
kilowatt-hours
feet
inches
miles
atmospheres
ft of water
pascals
pounds-force/sq in
ft/min
ft/s
kilometers/h
miles/h
pascal-seconds
barrels(oil)
cubic ft
gallons
liters
barrels(oil)/d
barrels(oil)
cubic meters
gallons
liters
cubic meters/s
barrels(oil)/d
liters/s
gallons/min
barrels(oil)
cubic ft
To convert from
cubic meters
cubic meters
cubic meters/s
darcies
dynes/sq cm
dynes
dynes
ergs
ergs
ergs
ergs
ergs/s
feet
feet
feet
feet of water
feet of water
feet of water
feet of water
feet/min
feet/min
feet/min
feet/s
feet/s
feet/s
feet/s2
foot-pounds
foot-pounds
foot-pounds
foot-pounds
foot-pounds
foot-pounds/min
foot-pounds/min
foot-pounds/min
foot-pounds/s
foot-pounds/s
foot-pounds/s
gallons
gallons
gallons
gallons
gallons/min
grams
horsepower
Multiply by
2.642 x 102
1.0 x 103
5.434 x 105
9.869233 x10-13
9.869233 x 10-7
1.0 x 10-5
2.248 x 10-6
9.486 x 10-11
7.376 x10-8
1.0 x 10-7
2.773 x 10-14
1.341 x 10-10
3.048 x 10-4
3.048 x 10-1
1.894 x 10-4
2.950 x 10-2
2.242
2.989 x 103
4.335 x 10-1
1.829 x 10-2
3.048 x 10-1
1.136 x 10-2
1.097
1.829 x 101
6.818 x 10-1
3.048 x 10-1
1.286 x 10-3
1.356 x 107
1.356
1.383 x 10-1
3.766 x 10-7
1.286 x 10-3
3.030 x 10-5
2.260 x 10-5
4.6263
1.818 x 10-3
1.356 x 10-3
2.381 x 10-2
1.337 x 10-1
3.785 x 10-3
3.785
2.228 x 10-3
2.205 x 10-3
4.244 x 101
27
To obtain
gallons
liters
barrels(oil)/d
sq meters
atmospheres
joules/meter (newtons)
pounds-force
Btu
ft-pounds
joules
kilowatt-hrs
horsepower
kilometers
meters
miles
atmospheres
cm of mercury
pascals
pounds-force/sq in
km/hr
meters/min
miles/h
km/h
meters/min
miles/h
meters/s2
Btu
ergs
joules
kg-m
kilowatt-h
Btu/min
horsepower
kilowatts
Btu/h
horsepower
kilowatts
barrels(oil)
cubic ft
cubic meters
liters
cu ft/s
pounds-mass
btu/min
To convert from
horsepower
horsepower
horsepower
horsepower-hours
horsepower-hours
horsepower-hours
inches
inches
joules
joules
joules/s
kilograms
kilograms/cu m
kilometers
kilometers
kilometers/h
kilometers/h
kilometers/h
kilowatts
kilowatt hour
kilowatt hour
kilowatt hour
liters
liters
liters
liters
liters/s
meters
meters
meters
meters
meters/s
meters/s
meters/s
meters/s
miles
miles
miles
miles/h
miles/h
miles/h
miles/h
newtons
pounds-force
Multiply by
3.3 x 104
5.50 x 102
7.457 x 10-1
2.547 x 103
1.98 x 106
2.684 x 106
2.540
2.540 x 10-2
9.486 x 10-4
7.376 x 10-1
5.6907 x 10-2
2.2046
6.243 x 10-2
3.281 x 103
6.214 x 10-1
5.468 x 101
9.113 x 10-1
6.214 x 10-1
1.341
3.413 x 103
2.655 x 106
3.6 x 106
1.0 x 103
3.531 x 10-2
1.0 x 10-3
2.642 x 10-1
1.5850 x 101
3.281
3.937 x 101
6.214 x 10-4
1.094
1.968 x 102
3.281
3.6
2.237
5.280 x 103
1.609
1.609 x 103
8.8 x 101
1.467
1.6093
2.682 x 101
1.0 x 105
4.4482 x 105
28
To obtain
ft-lb/min
ft-lb/s
kilowatts
Btu
ft-lb
joules
centimeters
meters
Btu
ft-pounds
Btu/min
pounds-mass
pounds-mass/cu ft
ft
miles
ft/min
ft/s
miles/h
horsepower
Btu
ft-lb
joules
cubic cm
cubic ft
cubic meters
gallons
gallons/min
ft
inches
miles
yards
ft/min
ft/s
km/h
miles/h
ft
kilometers
meters
ft/min.
ft/s
km/h
meters/min
dynes
dynes
To convert from
pounds-force
pounds-mass
pounds-mass
pounds-mass/cu ft
pounds-force/sq ft
pounds-force/sq ft
pounds-force/sq ft
pounds-force/sq in
pounds-force/sq in
pounds-force/sq in
radians
radians/s
revolutions/min
square cm
square feet
square feet
square feet
square meters
square meters
square meters
square miles
square miles
watts
watts
watts
watts
watts
watt hours
watt hours
Multiply by
4.4482
4.5359 x 102
4.5359 x 10-1
1.602 x 101
4.725 x 10-4
1.602 x 10-2
4.788 x 10
2.307
5.171
6.895 x 103
5.7296 x 101
9.5493
1.047 x 10-1
1.076 x 10-3
2.296 x 10-5
9.29 x 10-2
3.587 x 10-8
2.471 x 10-4
1.076 x 101
3.861 x 10-7
2.788 x 107
2.590 x 106
3.4129
1.0 x 107
4.427 x 101
7.376 x 10-1
1.341 x 10-3
3.413
2.656 x 103
29
To obtain
newtons
grams
kilograms
kg/cu m
atmospheres
ft of water
pascals
ft of water
cm of mercury
pascals
degrees
revolutions/min
radians/s
sq ft
acres
sq m
sq miles
acres
sq ft
sq miles
sq ft
sq meters
Btu/h
ergs/s
ft-lb/min
ft-lb/s
horsepower
Btu
ft-pounds
30